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Section 5.1 #54: Determine if the function is one-to-one. If it is, find a formula for the inverse function:
f(x)=(x+5)3.
Solution: First, the graph of this function is obtained by starting with the graph of y=x3 and then shifting it to the left by 5 units:
,
which passes the "horizontal line test" and hence is one-to-one. So finding the inverse is possible. We have
y=(x+5)3.
Solve for x by first taking the cube root to get
3√y=3√(x+5)3.
So,
3√y=x+5.
Therefore
3√y−5=x.
This means that f−1(y)=3√y−5.
(note: writing f−1(x)=3√x−5 is also ok -- you would arrive here if you do the "switch x and y" thing)
Section 5.2 #34: Sketch the graph of
y=2x−1−3.
Solution: This is the graph of y=2x shifted to the right by 1 and then shifted down by 3. Therefore we get:
Section 5.3 #10: Find log3(9).
Solution: The key is to recall that log3(3x)=x, and so we want to rewrite 9 as a power of 3. This yields
log3(9)=log3(32)=2.
Section 5.3 #33: Find ln(√e).
Solution: Recall that ln is simply loge, where e is the special number e=2.71…. Therefore we want to express √e as a power of e and then take the logarithm:
ln(√e)=ln(e12)=12.
Section 5.3 #46: Convert to an exponential equation:
t=log4(7).
Solution: We would like the logarithm on the right to disappear. Recall that 4log4(x)=x, and so we plug both sides of the equation into 4x to get
4t=4log4(7)
yielding
4t=7.
Section 5.3 #90: Sketch the graph of
y=ln(x+1).
Solution: This is the graph of y=ln(x) shifted to the left by 1 unit. Therefore,